大数乘法

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;


char s1[1005], s2[1005];
int n1[1005], n2[1005], ans[2005];

int main() {
    cin >> s1 >> s2;
    n1[0] = strlen(s1), n2[0] = strlen(s2);
    for (int i = 0, j = n1[0]; i < n1[0]; i++, j--) {
        n1[j] = s1[i] - '0';
    }
    for (int i = 0, j = n2[0]; i < n2[0]; i++, j--) {
        n2[j] = s2[i] - '0';
    }
    ans[0] = n1[0] + n2[0] - 1;
    for (int i = 1; i <= n1[0]; i++) {
        for (int j = 1; j <= n2[0]; j++) {
            ans[i + j -1] += n1[i] * n2[j];
        }
    }
    for (int i = 1; i <= ans[0]; i++) {
        if (ans[i] > 9) {
            ans[i + 1] += ans[i] / 10;
            ans[i] %= 10;
            if (i == ans[0]) {
                ans[0]++;
            }
        } else if (i == ans[0] && ans[i + 1] != 0) {
            ans[i]++;
        }
    }
    for (int i = ans[0]; i > 0; i--) {
        cout << ans[i];
    }
    cout << endl;
    return 0;
}

方法二

#include<iostream>
using namespace std;

long long ans[25][25];

int main() {
    long long fin = 1;
    for (int i = 40, j = 1; i > 20; i--, j++) {
        fin *= i;
        fin /= j;
    }
    cout << fin <<endl;
    return 0;
}